Anders Englöf Ytterström
6001a6bb59
Very intermediate pythonic solution, regex would have made the code more compact. But since 2023:01 decreased the regex courage, This code will do.
53 lines
1.2 KiB
Python
53 lines
1.2 KiB
Python
import re
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from math import prod
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from output import answer, puzzleinput
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from collections import defaultdict
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n = 2
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title = "Cube Conundrum"
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@answer(1, "Sum of all possible game IDs: {}")
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def part_1(data):
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return sum(
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[
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id + 1
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for id, game in enumerate(data.splitlines())
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if not sum(
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any(
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[
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c == "red" and int(n) > 12,
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c == "green" and int(n) > 13,
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c == "blue" and int(n) > 14,
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]
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)
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for n, c in re.findall(r"(\d+) (\w+)", game)
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)
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]
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)
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@answer(2, "Sum of all cube set powers: {}")
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def part_2(data):
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def power(game):
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seen = defaultdict(int)
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for n, c in re.findall(r"(\d+) (\w+)", game):
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seen[c] = max([seen[c], int(n)])
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return prod([seen["blue"], seen["red"], seen["green"]])
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return sum(power(game) for game in data.splitlines())
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@puzzleinput(n)
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def parse_input(data):
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return data
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if __name__ == "__main__":
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parsed = parse_input()
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a = part_1(parsed)
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b = part_2(parsed)
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assert a == 2439
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assert b == 63711
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